Prove that the radii of the circles x2 + y2 = 1, x2 + y2 – 2x – 6y – 6 = 0 and x2 + y2 – 4x – 12y – 9 = 0 are in A. P.
Given circles are:
(i) x2 + y2 - 1 = 0
(ii) x2 + y2 - 2x - 6y - 6 = 0
(iii) x2 + y2 - 4x - 12y - 9 = 0
We know that for a circle x2 + y2 + 2ax + 2by + c = 0 - - (1)
⇒ Centre = (- a, - b)
⇒ Radius = 
Comparing (i) with (1) we get,
⇒ Radius (r1) = 
⇒ r1 = 1
Comparing (ii) with (1) we get,
⇒ Radius (r2) = 
⇒ 
⇒ r2 = √16
⇒ r2 = 4
Comparing (iii) with (1) we get,
⇒ Radius (r3) = 
⇒ 
⇒ r3 = √49
⇒ r3 = 7
We need to show r1,r2,r3 are in A.P.
We know that for three numbers a, b, c to be in A.P. The condition to be satisfied is:
⇒ 
⇒ 
⇒ 
⇒ 
⇒ b = 4
⇒ b = r2
∴ The radii r1,r2,r3 are in A.P.
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